Problem

有以下操作

  1. 插入x数
  2. 删除x数(若有多个相同的数,因只删除一个)
  3. 查询x数的排名(若有多个相同的数,因输出最小的排名)
  4. 查询排名为x的数
  5. 求x的前驱(前驱定义为小于x,且最大的数)
  6. 求x的后继(后继定义为大于x,且最小的数)

Solution

裸的Splay:
Rotate旋转操作
Splay伸展操作

Notice

注意,删除的时候不是把这个元素都删除,而是删除一个!!!

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int point = 0, root, pre, suf, ans;
struct node
{
	int val[N + 5], count[N + 5], num[N + 5], son[2][N + 5], parent[N + 5];
	inline void up(int u)
	{
		count[u] = count[son[0][u]] + count[son[1][u]] + num[u];
	}
	
	void Rotate(int x, int &rt)
	{
		int y = parent[x], z = parent[y];
		int l = (son[1][y] == x), r = 1 - l;
		if (y == rt) rt = x;
		else if (son[0][z] == y) son[0][z] = x;
		else son[1][z] = x;
		parent[x] = z;
		parent[son[r][x]] = y, son[l][y] = son[r][x];
		parent[y] = x, son[r][x] = y;
		up(y);
		up(x);
	}
	
	void Splay(int x, int &rt)
	{
		while (x != rt)
		{
			int y = parent[x], z = parent[y];
			if (y != rt)
			{
				if ((son[0][z] == y) ^ (son[0][y] == x))
					Rotate(x, rt);
				else Rotate(y, rt);
			}
			Rotate(x, rt);
		}
	}
	
	void Insert(int &u, int x, int last)
	{
		if (u == 0)
		{
			u = ++point;
			val[u] = x, parent[u] = last, num[u] = count[u] = 1;
			Splay(u, root);
		}
		else
		{
			if (x > val[u]) Insert(son[1][u], x, u);
			else if (x < val[u]) Insert(son[0][u], x, u);
			else if (x == val[u]) num[u]++, count[u]++, Splay(u, root);
		}
	}
	
	void Delete(int x)
	{
		Splay(x, root);
		if (num[x] > 1) 
		{
			num[x]--, count[x]--;
			return;
		}
		if (son[0][x] * son[1][x] == 0) root = son[0][x] + son[1][x];
		else 
		{
			int t = son[1][x];
			while (son[0][t] != 0) t = son[0][t];
			Splay(t, root);
			son[0][t] = son[0][x], parent[son[0][x]] = t;
			up(t);
		}
		parent[root] = 0;
	}
	
	void Find_pre(int u, int x)
	{
	    if (u == 0) return;
	    if (x > val[u])
	    {
	        pre = u;
	        Find_pre(son[1][u], x);
	        ans += count[son[0][u]] + num[u];
	    }
	    else Find_pre(son[0][u], x);
	}
	
	void Find_suf(int u,int x)
	{
	    if (u == 0) return;
	    if (x < val[u])
	    {
	        suf = u;
	        Find_suf(son[0][u], x);
	    }
	    else Find_suf(son[1][u], x);
	}
	
	int Find_num(int u, int x)
	{
		if (x <= count[son[0][u]]) return Find_num(son[0][u], x);
		if (x > count[son[0][u]] + num[u]) return Find_num(son[1][u], x - count[son[0][u]] - num[u]);
		return val[u];
	}
	
	int Find_rank(int x)
	{
		ans = 0;
		Find_pre(root, x);
		return ans + 1;
	}
	
	int Find_id(int u, int x)
	{
		if (x == val[u]) return u;
		if (x > val[u]) return Find_id(son[1][u], x);
		if (x < val[u]) return Find_id(son[0][u], x);
	}
}splay_tree;
int sqz()
{
	int n = read();
	root = 0;
	rep(i, 1, n)
	{
		int op = read(), t = read();
		switch (op)
		{
			case 1:
			{
				splay_tree.Insert(root, t, 0);
				break;
			}
			case 2:
			{
				int x = splay_tree.Find_id(root, t);
				splay_tree.Delete(x);
				break;
			}
			case 3:
			{
				printf("%d\n", splay_tree.Find_rank(t));
				break;
			}
			case 4:
			{
				printf("%d\n", splay_tree.Find_num(root, t));
				break;
			}
			case 5:
			{
				splay_tree.Find_pre(root, t);
				printf("%d\n", splay_tree.val[pre]);
				break;
			}
			case 6:
			{
				splay_tree.Find_suf(root, t);
				printf("%d\n", splay_tree.val[suf]);
				break;
			}
		}
	}
	return 0;
}
posted on 2017-09-29 20:25  WizardCowboy  阅读(115)  评论(0编辑  收藏  举报